MATHEMATICS
Closed G 6 subsets of supercompact Hausdorff spaces
by Jan van Mill and Charles F. Mills" J. van ll!ill ; Wiskundig Sem inarium, Free University, Amsterdam Oh, F. Mil18; Dept. of Mathematics, Un iversity of Wisconsin , Madison ,
Wisconsin 53706
Communicated by Prof. W . T. van Est at the meeting of June 17, 1978
ABSTRACT
We give examples of compact Hausdorff spaces which aro not embed dable as closed G" subsets in a supercompact Hausdorff space. K oy words and phrases: supereompaet, linked system, Cantor t ree, density. A.'I\1S(MOS) subject olaasification (1970) : 54D35. INTRODUCTION
A supercompact space is a space which has a binary subbase for its closed subsets, where a collection of subsets !7 of a set X is called binary provided that for all J( C !7 with r. Jt = 0 there are M 0, M 1 E Jt with M 0 r. M 1 = 0. By Alexander's subbase lemma, every supercompact space is compact. The class of supercompact spaces was introduced by de Groot [9]. Many spaces are supercompact, for example all compact metric spaces, cf. Strok & Szymanski [I4J (elementary proofs of this fact were recently found by van Douwen [6J and Mills [I2J). The first examples of nonsupercompact compact Hausdorff sp aces were found by Bell [IJ. At the moment there is a variety of nonsupercompact compact Hausdorff spaces (cf. Bell [I J, [2], van Douwen & van Mill [7J, van Mill [IlJ, Bell & van Mill [4J). • The first author is supported ty the Netherlands Organization for the Advancement of Pure Research (Z.W.O.); Juliana van Stolberglaan 148, 's-Gravenhage, the Netherlands.
155
Recently, Bell [3] showed that the one point compactification of the Cantor tree 312 u"'2 (cf. Rudin [13]) can be embedded as a closed 06 subset of a supercompact Hausdorff space. Since the one point compaetification of the Cantor tree is not supercompact (cf. van Douwen & van Mill [7]) this yields an example of a nonsupercompact closed 0 6 in a supercompact Hausdorff space. This suggests the question whether every compact Hausdorff space can be embedded as a 06 subset in a supercompact Hausdorff space. The answer to this question is in the negative. 0.1. THEOREM: Let X be a Hausdorff continuous image 01 a closed 0 6 subset 01 a supercompact Hausdorff space, and let K be a closed subset 01 X such that IKI > 2"'. Then at least one point 01 K is the limit 01 a nontrivial convergent sequence in X (not necessarily in K). This theorem is a consequence of a result in van Douwen & van Mill [7]. As a corollary, if fJX is a continuous image of a closed 06 subset of a supercompact Hausdorff space then X is pseudocompact. Also, under Martins axiom (MA), every infinite Hausdorff continuous image of a closed 0 6 subset of a supercompact Hausdorff space contains a nontrivial convergent sequence. Since the one point compactification of the Cantor tree is a compactification of co with the one point compactification of a discrete space as remainder, Bell's [3] result suggests the question whether every compactification of co with the one point compactification of a discrete space as remainder can be embedded as a G« sunset of a supercompact Hausdorff space. The answer to this question is in the negative. For every (faithfully indexed) almost disjoint family .A' = {M",I IX E ,,} of infinite subsets of co define X.,I( to be the space with underlying set the disjoint union of " and co and with topology generated by the collection
{{IX}
U
(M",-n)IIX E x, n E co}
U
{{n}ln E co}.
Notice that X.,I( is separable and that every subspace of X.,I( is locally compact and first countable. Also, the Cantor tree 312 U"'2 is homeomorphic to some X"If. We will prove the following theorem: 0.2. THEOREM: Let .A' be a maximal uncountable almost disjoint collection 01 infinite subsets 01 co. Then any compactification 01 X"H is not the continuous image 01 a closed G6 subset 01 a supercompact Hausdorff space.
1.
THEOREM
0.1;
PROOF AND CONSEQUENCES
1.1. PROOF OF THEOREM 0.1: Indeed, let Y be a supercompact Hausdorff space, let X and K be as in Theorem 0.1 and let Z be a closed 06 in Y which is mapped by 1 onto X. Write Z= nn~", Un, where the Un'S are open subsets of Y. It is easily verified that a space has a binary 156
subbase if and only if it has a binary subbase closed under arbitrary intersections. Let g be a binary subbase for Y which is closed under arbitrary intersections. For each nEW let / n be a finite subcollection of g such that Z C U /n C Un. For each z E Z and nEW take Fn(z) E /n containing z. In addition, for each ZEZ define F(z):= nnE",Fn{z). Then F{z) E g for each Z E Z, hence F(z) is supercompact, UZEZ F(z) = Z and the collection {F(z)/z E Z} has cardinality at most 2"'. Since IKI > 2'" there is a z E Z and a countably infinite subset E C K such that E C f[F(z)]. By a theorem in van Douwen & van Mill [7] it follows that at least one cluster point of E is the limit of a nontrivial convergent sequence in f[F{z)]. This completes the proof. 0 1.2. COROLLARY: Suppose that pX is a continuous image of a closed G6 subset of a supercompact Hausdorff space. Then X is pseudocompaci. PROOF: Assume that X is not pseudocompact. Then we may assume that we X and that t» is C-embedded in X (cf. Gillman & Jerison [8]). Then fJw - w C pX - X and since IfJw - wi = 22'" (cf. Gillman & Jerison [8]) by Theorem 0.1 there is an x E pw - co which is the limit of a nontrivial convergent sequence in fJX. It is easily seen that this is impossible. 0
Recall that Martin's axiom (MA) states that no compact ccc Hausdorff space is the union of less than 2'" nowhere dense sets (cf. Martin & Solovay [10]). It is known (cf. Booth [5]) that MA implies P{2"'), i.e. the statement that for every collection .SiI of fewer than 2'" subsets of w such that each finite su bcollection of .SiI has infinite intersection there is an infinite Few such that F - A is finite for all A E.SiI. It is easily seen that P( 2"') implies that Bco - w is not the union of 2'" nowhere dense sets. This implies that, under P{2"'), every compactification yw of w with the property that no sequence in w converges has cardinality greater than 2"'. For let yw be such a compactification of wand let f: pw ~ yw be the unique continuous surjection which extends the identity on w. Now the fact that no sequence in w converges implies that f-l(x) is nowhere dense in fJw - w for all xEyw-W. Hence P(2"') implies that lyw-wl>2"'. 1.3. COROLLARY (P(2"'»): Let X be a Hausdorff continuous image of a closed G6 subset of a supercompact Hausdorff space. If X is infinite then X contains a nontrivial convergent sequence. If jX!>2'" then this follows from Theorem 0.1. On the other hand, if IXI <: 2'" then this follows from P( 2"'). 0 PROOF:
1.4.
QUESTION:
Is Corollary 1.3 true in ZFC 1
157
2.
PROOF OF THEOREM
0.2
Recall that a family of subsets .91 of w is called almost disjoint provided that A ('\ B is finite for all distinct A, B E.9I. It is known that there is an almost disjoint family .91 C 9i'(w) of cardinality 2'" (cf. Gillman & Jerison [8]). We need the following lemma. 2.1. LEMMA: Let {A..llX E ,,} be an uncountable (faithfully indexed) maximal almost disjoint jamily oj infinite subsets oj os, Ij {Pn : W ~ mn} is a sequence oj partitions oj W into finitely many sets, then there is an j E "'w such that
I nnE'" {lXIIA.. ('\ nhn P;l(f(i)) I= w}1 > WI· PROOF:
(1)
We choose j(n)
E
mn by induction so that
for every finite F C" we have that p ,- l(f(i )) - Ui,FAi!=w. I
ntu
Indeed, suppose that {j(i)/i E n} have been defined such that (1) is satisfied. If n,.-, 0, then define j(O) to be an arbitrary element of mo such that for every finite F C" we have that /p;l(f(O)) - U11FA j i = w. It is clear that this is possible since mo is finite and " is infinite. If n =;6 0 then define M n - 1 : = n,ln-1 P,-l(f(i))
and notice that
is an uncountable maximal almost disjoint family of infinite subsets of M n-l. Since P n ~ M n-l is a partition of .iJ:fn- 1 and since M n - l is infinite by induction hypothesis there is an m E mn such that J(Pn ~Mn-l)-l(m)-
U
fl=w
for every finite subcollection fC.9I'. Now define j(n):=m; then it is clear that (1) is satisfied. Suppose that there are only eountably many lX, say {lXmlm E w}, such that for all n, m e to we have that IA"'m ('\ P;l(f(i))1 =W. Then we may pick, by (1), distinct pn E W such that
ntu
pn Entin P-I(f(i)) -
U}n A",;
(n E w).
Define A: = {Pnln E w}. There are two cases: suppose first that A E {A a IlX E x]. Then, since /A ('\ n',n Pi l(f(i))l =W for all nEW we have that A =A"'m for some m, which is impossible by definition of the Pn's. Therefore A ¢ {A ..llX c: "J. By maximality we can find a fJ E" such that IA fJ ('\ A I= w. Since
IA 158
n,u P.-1(j(i)) I
we conclude that IA/l n n(en P;l(f(i))1 =co for all n E co, so f3 = IXm for some m. But since IA n A ",,,I < co for all n E co we have a contradiction. D We now can prove the main result in this section. 2.2. PROOF OF THEOREM 0.2: List J( as {M",IIXE"}. Assume that Y is a supercompaot Hausdorff space, that Z C Y is a closed Gd and that g: Z ---+ yX.,I( is a continuous surjection from Z onto the compaotifloation yX.,I( of X.,I(. Let .9 be a binary subbase for Y which is closed under arbitrary intersections. Let {Ullin E co} be a sequence of open subsets of Y whose intersection is Z. Since Un-g-l(n) is a neighborhood of Z-g-l(n) and since Z-g-l(n) is closed in Y, we can find 8&, ... ,8::',,-2 E.9 such that UII-g-l(n)'JS&u ... U8::',,_2'JZ-g-1(n). For each nEco pick d ll E Z such that g(d ll ) =n. Define D: = {dliinE co}. Take P II : co ---+ mil to be a partition refining {87 n Dlj Emil-I} U {d(i)li En}, in such a way that P;;1(j)C87flD for eachjEmll-I and p;;l({mn-I})={d(i)liEn}. For each IX E" let A",: = {d(n) In EM",}. Now pick f as in Lemma 2.1. We then have, by the compactness of Z, that g( nUa>8j("1 'J nlieCD {IXIIA", fl n(ell Pi"l(f(i»1 = co}.
Let 8:= nlle",8j(,,). Notice that 8CZ-g-l(co) and in addition that 8 is uncountable by Lemma 2.1. For each IX E" the set g-l(M", U {IX}) is open and closed in Z. Hence we may take an open set V", C Y(IX E ,,) such that ely (V",)
fl Z= V .. fl Z=g-l(M.. U
{IX}).
Notice that for distinct IX, f3 E" we have that V.. U V/l C g-l(co) U (Y -Z). Set H = nll.a> {IXIIA.. fl ncen Pi"l(f(i»1 = co}. For each IX E H let J a be a finite subcollection of.9 such that g-l(M.. U {IX}) C U J .. C V... Since / .. is finite we may take 8",E J a such that IA", fl ncell Pi"l(f(i)) fl 8 ..\= co for all n E co. Since D is countable and H is uncountable there exist distinct IX, f3 E H such that 8", fl 8/l=F0. It is clear that 8 .. fl 8/l fl 8=8.. fl 8 p fl nU",8j("1 C V",
fl
V/l
fl (Z-g-l(co»
= 0.
Therefore, since .9 is binary and since 8 .. fl 8/l i= 0, we may assume, without loss of generality, that there is an no E co such that 8 .. fl 8~r:.u) = 0. However, since P;;;/(f(no») C 8~~ )and since IA.. fl ncCllop;l(f(i» fl 8 ..1=co this is a. contradiction. 0 D Gd'S IN SUPERCOMPACT HAUSDORFF SPACES In this section we show that if Z is a closed Gd in a supercompact Hausdorff space X then d(Z) < 2"'d(X). 3.
DENSITY OF CLOSED
159
Recall that the density d(X) of a topological space X is the least cardinal x for which there is a dense subset of cardinality x. If 1/ is a binary subbase for X then for all A C X we define I(A) ex by I(A):=
n
{S
E
I/IA C S}.
Notice that clx (A) C I(A), since each element of 1/ is closed, that I(I(A))=I(A) and that I(A) C I(B) if A C B C X. The following lemma was proved in van Douwen & van Mill [7]. For the sake of completeness we will give its proof here also. 3.1. LEMMA: Let 1/ be a binary subbase for the supercompact Hausdorff space X. Let p EX. If U is a neighborhood of p and if A is a subset of X with p E clx (A), then there is a subset B C A with p E clx (B) and I(B) C U.
Since X is regular, p has a neighborhood V such that denote the collection of finite intersections of elements from 1/. Choose a finite ,/ C f such that cl.x (V) C u ,/ CU. Now ,/ is finite, and A n V C U ,/, and p E clx (A n V); hence there is an S E ,/withp E clx (A n V n S). LetB: =A n V n S. Thenp E clz (B), and Be A, and I(B) esc U , / C U. 0 PROOF:
p E clz (V) CU. Let f
We now prove the main result in this section. 3.2. THEOREM: Let f/ be a binary subbase for the Hausdorff space X. Then d(S)<,d(X) for all S E s: Let D be a dense subset of X and choose S E f/. For each choose a point e(d) E nSE.9' I ({d, s}) n S . Notice that this is possible since f/ is binary. We claim that E:={e(d)jdED} is dense in S. Indeed, take XES and let U be any neighborhood cf x. By Lemma 3.1 there is a subset BCD such that x is in the closure of Band I(B) CU. Choose do E D arbitrarily. Then e(do) E I({d o, s}) n S C I({do, x}) n S C I(B) n S C ti r, S. This completes the proof. 0 PROOF:
dE D
n8ls
3.3. COROLLARY: Let Z be a closed dorff space X. Then d(Z) 0;;;; 2
G~
subset in a supercompact Haus-
QJd(X).
Let 1/ be a binary subbase for X which is closed under arbitrary intersections. As in the proof of Theorem 0.1, Z is the union of a family of at most 2 subsets of 1/. Hence Theorem 3.2 implies that d(Z) <: 2 PROOF:
QJ
QJd(X).
o 160
4.
OPEN QUESTIONS
The results derived in this note suggest many questions. As noted in the introduction Bell [3] has shown that a closed G~ subset of a supercompact Hausdorff space need not be supercompact. This suggests the following question. 4.1. QUESTION: Suppose that Z is a closed Hausdorff space X. Is cmpn(Z) finite?
G~
in a supercompact
(Recall that for compact Hausdorff spaces X, cmpn(X) is the least integer k for which there is a closed subbase [/ for X such that if JI C [/ with f"'I JI = 0 then there is a subset of JI of cardinality k which has an empty intersection; cmpn(X) = 00 if such an integer does not exist (cf. Bell & van Mill [4]). It is known, cf. [4], that for every k;» I there is a compact Hausdorff space X lc for which cmpn(Xlc)=k; in addition cmpn(fJw) = 00 ). Related to this question is the following one: 4.2. QUESTION: Suppose that fJX is a continuous image of a closed G~ of a compact Hausdorff space Y with cmpn(Y) <00. Is X pseudocompact? 4.3. QUESTION: Let X be an infinite compact Hausdorff space for which cmpn(X) <00. Does X contain a copy of w which is not C*-embedded in X? a nontrivial convergent sequence?
In section 2 we gave an example of a compact Hausdorff space X which is the union of three metrizable subspaces and which is not embeddable as a G~ subset in a supercompact Hausdorff space. This suggests the following question. 4.4. QUESTION: Let X be a compact Hausdorff space which is the union of two metrizable subspaces. Can X be embedded as a G~ subset in a supercompact H ausdorU space?
REFERENCES
1. Bell, M. G. - Not all compact Hausdorff spaces are supercompact, Gen. Top. Appl. 8, 151-155 (1978). 2. Bell, M. G. - A cellular constraint in supercompact Hausdorff spaces, Canadian J. Math. 30, 1144-1151 (1978). 3. Bell, M. G. - A first countable supercompact Hausdorff space with a closed G~ non-supercompact subspace (to appear in Colloq. Math.). 4. Bell, M. G. & J. van Mill - The compactness number of a compact topological space (to appear in Fund. Math.). 5. Booth, D. - ffitrafilters on a countable set, Ann. Math. Logic 2, 1-24 (1970). 6. Douwen, E. K. van - Special bases for compact metrizable spaces, (to appear in Fund. Math.).
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7. Douwen E. K. van & J. van Mill - Supercompact spaces, (to appear in Gen. Top. Appl.), 8. Gillman,L. &M. Jerison-Rings ofcontinuous functions, Princetion, N.J. (1960). 9. Groot, J. de - Supercompactness and superextensions, in: Contributions to extension theory of topological structures, Symp. Berlin 1967, Deutscher Verlag Wiss., Berlin 89-90 (1969). 10. Martin, D. A. & R. M. Solovay - Internal Cohen extensions, Ann. Ma.th. Logic 2, 143-178 (1970). 11. Mill, J. van - A countable space no oompaetifleabion of which is supercompact, Bull. l'aoad. Pol. ser., 25 (1977) 1129-1132. 12. Mills, C. F. - A simpler proof that compact metric spaces are supercompaet., (to appear in Proc, Amer, Math. Soc.), 13. Rudin, M. E. - Lectures on set theoretic topology, Regional Conf. Ser, in Math., no. 23, Am. Math. Soc., Providence, RI (1975). 14. Strok, M. & A. Szymanski - Compact metric spaces have binary bases, Fund. Math. 89, 81-91 (1975).
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